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                    <h1 class="description center-align post-title">深度学习——Backpropagation</h1>
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                <p>本文基于<a target="_blank" rel="noopener" href="https://github.com/Sakura-gh">Sakura-gh</a>大佬的机器学习笔记修改，仅作为学习资料备用，如有侵权，联系作者。</p>
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<h1 id="Backpropagation"><a href="#Backpropagation" class="headerlink" title="Backpropagation"></a>Backpropagation</h1><blockquote>
<p>Backpropagation(反向传播)，就是告诉我们用gradient descent来train一个neural network的时候该怎么做，它只是求微分的一种方法，而不是一种新的算法</p>
</blockquote>
<h4 id="Gradient-Descent"><a href="#Gradient-Descent" class="headerlink" title="Gradient Descent"></a>Gradient Descent</h4><p>gradient descent的使用方法，跟前面讲到的linear Regression或者是Logistic Regression是一模一样的，唯一的区别就在于当它用在neural network的时候，network parameters $\theta=w_1,w_2,…,b_1,b_2,…$里面可能会有将近million个参数</p>
<p>所以现在最大的困难是，如何有效地把这个近百万维的vector给计算出来，这就是Backpropagation要做的事情，所以<strong>Backpropagation并不是一个和gradient descent不同的training的方法，它就是gradient descent，它只是一个比较有效率的算法</strong>，让你在计算这个gradient的vector的时候更有效率</p>
<h4 id="Chain-Rule"><a href="#Chain-Rule" class="headerlink" title="Chain Rule"></a>Chain Rule</h4><p>Backpropagation里面并没有什么高深的数学，你唯一需要记得的就只有Chain Rule(链式法则)</p>
<p>对整个neural network，我们定义了一个loss function：$L(\theta)=\sum\limits_{n=1}^N l^n(\theta)$，它等于所有training data的loss之和</p>
<center><img src="https://gitee.com/Sakura-gh/ML-notes/raw/master/img/bp-loss.png" width="50%;"></center>
我们把training data里任意一个样本点$x^n$代到neural network里面，它会output一个$y^n$，我们把这个output跟样本点本身的label标注的target $\hat{y}^n$作cross entropy，这个**交叉熵定义了output $y^n$和target $\hat{y}^n$之间的距离$l^n(\theta)$**，如果cross entropy比较大的话，说明output和target之间距离很远，这个network的parameter的loss是比较大的，反之则说明这组parameter是比较好的

然后summation over所有training data的cross entropy $l^n(\theta)$，得到total loss $L(\theta)$，这就是我们的loss function，用这个$L(\theta)$对某一个参数w做偏微分，表达式如下：
$$
\frac{\partial L(\theta)}{\partial w}=\sum\limits_{n=1}^N\frac{\partial l^n(\theta)}{\partial w}
$$
这个表达式告诉我们，只需要考虑如何计算对某一笔data的$\frac{\partial l^n(\theta)}{\partial w}$，再将所有training data的cross entropy对参数w的偏微分累计求和，就可以把total loss对某一个参数w的偏微分给计算出来

我们先考虑某一个neuron，先拿出上图中被红色三角形圈住的neuron，假设只有两个input $x_1,x_2$，通过这个neuron，我们先得到$z=b+w_1 x_1+w_2 x_2$，然后经过activation function从这个neuron中output出来，作为后续neuron的input，再经过了非常非常多的事情以后，会得到最终的output $y_1,y_2$

现在的问题是这样：$\frac{\partial l}{\partial w}$该怎么算？按照chain rule，可以把它拆分成两项，$\frac{\partial l}{\partial w}=\frac{\partial z}{\partial w} \frac{\partial l}{\partial z}$，这两项分别去把它计算出来。前面这一项是比较简单的，后面这一项是比较复杂的

计算前面这一项$\frac{\partial z}{\partial w}$的这个process，我们称之为Forward pass；而计算后面这项$\frac{\partial l}{\partial z}$的process，我们称之为Backward pass

<center><img src="https://gitee.com/Sakura-gh/ML-notes/raw/master/img/bp-forward-backward.png" width="50%;"></center>
#### Forward pass

先考虑$\frac{\partial z}{\partial w}$这一项，完全可以秒算出来，$\frac{\partial z}{\partial w_1}=x_1 ,\ \frac{\partial z}{\partial w_2}=x_2$

它的规律是这样的：==**求$\frac{\partial z}{\partial w}$，就是看w前面连接的input是什么，那微分后的$\frac{\partial z}{\partial w}$值就是什么**==，因此只要计算出neural network里面每一个neuron的output就可以知道任意的z对w的偏微分

- 比如input layer作为neuron的输入时，$w_1$前面连接的是$x_1$，所以微分值就是$x_1$；$w_2$前面连接的是$x_2$，所以微分值就是$x_2$
- 比如hidden layer作为neuron的输入时，那该neuron的input就是前一层neuron的output，于是$\frac{\partial z}{\partial w}$的值就是前一层的z经过activation function之后输出的值(下图中的数据是假定activation function为sigmoid function得到的)

<center><img src="https://gitee.com/Sakura-gh/ML-notes/raw/master/img/forward-pass.png" width="50%;"></center>
#### Backward pass

再考虑$\frac{\partial l}{\partial z}$这一项，它是比较复杂的，这里我们依旧假设activation function是sigmoid function

##### 公式推导

我们的z通过activation function得到a，这个neuron的output是$a=\sigma(z)$，接下来这个a会乘上某一个weight $w_3$，再加上其它一大堆的value得到$z'$，它是下一个neuron activation function的input，然后a又会乘上另一个weight $w_4$，再加上其它一堆value得到$z''$，后面还会发生很多很多其他事情，不过这里我们就只先考虑下一步会发生什么事情：
$$
\frac{\partial l}{\partial z}=\frac{\partial a}{\partial z} \frac{\partial l}{\partial a}
$$
这里的$\frac{\partial a}{\partial z}$实际上就是activation function的微分(在这里就是sigmoid function的微分)，接下来的问题是$\frac{\partial l}{\partial a}$应该长什么样子呢？a会影响$z'$和$z''$，而$z'$和$z''$会影响$l$，所以通过chain rule可以得到
$$
\frac{\partial l}{\partial a}=\frac{\partial z'}{\partial a} \frac{\partial l}{\partial z'}+\frac{\partial z''}{\partial a} \frac{\partial l}{\partial z''}
$$
这里的$\frac{\partial z'}{\partial a}=w_3$，$\frac{\partial z''}{\partial a}=w_4$，那$\frac{\partial l}{\partial z'}$和$\frac{\partial l}{\partial z''}$又该怎么算呢？这里先假设我们已经通过某种方法把$\frac{\partial l}{\partial z'}$和$\frac{\partial l}{\partial z''}$这两项给算出来了，然后回过头去就可以把$\frac{\partial l}{\partial z}$给轻易地算出来
$$
\frac{\partial l}{\partial z}=\frac{\partial a}{\partial z} \frac{\partial l}{\partial a}=\sigma'(z)[w_3 \frac{\partial l}{\partial z'}+w_4 \frac{\partial l}{\partial z''}]
$$

<center><img src="https://gitee.com/Sakura-gh/ML-notes/raw/master/img/backward-pass.png" width="50%;"></center>
##### 另一个观点

这个式子还是蛮简单的，然后，我们可以从另外一个观点来看待这个式子

你可以想象说，现在有另外一个neuron，它不在我们原来的network里面，在下图中它被画成三角形，这个neuron的input就是$\frac{\partial l}{\partial z'}$和$\frac{\partial l}{\partial z''}$，那input $\frac{\partial l}{\partial z'}$就乘上$w_3$，input $\frac{\partial l}{\partial z''}$就乘上$w_4$，它们两个相加再乘上activation function的微分 $\sigma'(z)$，就可以得到output $\frac{\partial l}{\partial z}$

<center><img src="https://gitee.com/Sakura-gh/ML-notes/raw/master/img/backward-neuron.png" width="50%;"></center>
这张图描述了一个新的“neuron”，它的含义跟图下方的表达式是一模一样的，作这张图的目的是为了方便理解

值得注意的是，这里的$\sigma'(z)$是一个constant常数，它并不是一个function，因为z其实在计算forward pass的时候就已经被决定好了，z是一个固定的值

所以这个neuron其实跟我们之前看到的sigmoid function是不一样的，它并不是把input通过一个non-linear进行转换，而是直接把input乘上一个constant $\sigma'(z)$，就得到了output，因此这个neuron被画成三角形，代表它跟我们之前看到的圆形的neuron的运作方式是不一样的，它是直接乘上一个constant(这里的三角形有点像电路里的运算放大器op-amp，它也是乘上一个constant)

##### 两种情况

ok，现在我们最后需要解决的问题是，怎么计算$\frac{\partial l}{\partial z'}$和$\frac{\partial l}{\partial z''}$这两项，假设有两个不同的case：

###### case 1：Output Layer

假设蓝色的这个neuron已经是hidden layer的最后一层了，也就是说连接在$z'$和$z''$后的这两个红色的neuron已经是output layer，它的output就已经是整个network的output了，这个时候计算就比较简单
$$
\frac{\partial l}{\partial z'}=\frac{\partial y_1}{\partial z'} \frac{\partial l}{\partial y_1}
$$
其中$\frac{\partial y_1}{\partial z'}$就是output layer的activation function (softmax) 对$z'$的偏微分

而$\frac{\partial l}{\partial y_1}$就是loss对$y_1$的偏微分，它取决于你的loss function是怎么定义的，也就是你的output和target之间是怎么evaluate的，你可以用cross entropy，也可以用mean square error，用不同的定义，$\frac{\partial l}{\partial y_1}$的值就不一样

这个时候，你就已经可以把$l$对$w_1$和$w_2$的偏微分$\frac{\partial l}{\partial w_1}$、$\frac{\partial l}{\partial w_2}$算出来了

<center><img src="https://gitee.com/Sakura-gh/ML-notes/raw/master/img/bp-output-layer.png" width="50%;"></center>
###### Case 2：Not Output Layer

假设现在红色的neuron并不是整个network的output，那$z'$经过红色neuron的activation function得到$a'$，然后output $a'$和$w_5$、$w_6$相乘并加上一堆其他东西分别得到$z_a$和$z_b$，如下图所示

<center><img src="https://gitee.com/Sakura-gh/ML-notes/raw/master/img/not-output-layer.png" width="50%;"></center>
根据之前的推导证明类比，如果知道$\frac{\partial l}{\partial z_a}$和$\frac{\partial l}{\partial z_b}$，我们就可以计算$\frac{\partial l}{\partial z'}$，如下图所示，借助运算放大器的辅助理解，将$\frac{\partial l}{\partial z_a}$乘上$w_5$和$\frac{\partial l}{\partial z_b}$乘上$w_6$的值加起来再通过op-amp，乘上放大系数$\sigma'(z')$，就可以得到output $\frac{\partial l}{\partial z'}$
$$
\frac{\partial l}{\partial z'}=\sigma'(z')[w_5 \frac{\partial l}{\partial z_a} + w_6 \frac{\partial l}{\partial z_b}]
$$

<center><img src="https://gitee.com/Sakura-gh/ML-notes/raw/master/img/bp-not-output-layer.png" width="50%;"></center>
知道$z'$和$z''$就可以知道$z$，知道$z_a$和$z_b$就可以知道$z'$，...... ，现在这个过程就可以反复进行下去，直到找到output layer，我们可以算出确切的值，然后再一层一层反推回去

你可能会想说，这个方法听起来挺让人崩溃的，每次要算一个微分的值，都要一路往后走，一直走到network的output，如果写成表达式的话，一层一层往后展开，感觉会是一个很可怕的式子，但是！实际上并不是这个样子做的

你只要换一个方向，从output layer的$\frac{\partial l}{\partial z}$开始算，你就会发现它的运算量跟原来的network的Feedforward path其实是一样的

假设现在有6个neuron，每一个neuron的activation function的input分别是$z_1$、$z_2$、$z_3$、$z_4$、$z_5$、$z_6$，我们要计算$l$对这些$z$的偏微分，按照原来的思路，我们想要知道$z_1$的偏微分，就要去算$z_3$和$z_4$的偏微分，想要知道$z_3$和$z_4$的偏微分，就又要去计算两遍$z_5$和$z_6$的偏微分，因此如果我们是从$z_1$、$z_2$的偏微分开始算，那就没有效率

<center><img src="https://gitee.com/Sakura-gh/ML-notes/raw/master/img/input-z.png" width="50%;"></center>
但是，如果你反过来先去计算$z_5$和$z_6$的偏微分的话，这个process，就突然之间变得有效率起来了，我们先去计算$\frac{\partial l}{\partial z_5}$和$\frac{\partial l}{\partial z_6}$，然后就可以算出$\frac{\partial l}{\partial z_3}$和$\frac{\partial l}{\partial z_4}$，最后就可以算出$\frac{\partial l}{\partial z_1}$和$\frac{\partial l}{\partial z_2}$，而这一整个过程，就可以转化为op-amp运算放大器的那张图

<center><img src="https://gitee.com/Sakura-gh/ML-notes/raw/master/img/bp-op-amp.png" width="50%;"></center>
这里每一个op-amp的放大系数就是$\sigma'(z_1)$、$\sigma'(z_2)$、$\sigma'(z_3)$、$\sigma'(z_4)$，所以整一个流程就是，先快速地计算出$\frac{\partial l}{\partial z_5}$和$\frac{\partial l}{\partial z_6}$，然后再把这两个偏微分的值乘上路径上的weight汇集到neuron上面，再通过op-amp的放大，就可以得到$\frac{\partial l}{\partial z_3}$和$\frac{\partial l}{\partial z_4}$这两个偏微分的值，再让它们乘上一些weight，并且通过一个op-amp，就得到$\frac{\partial l}{\partial z_1}$和$\frac{\partial l}{\partial z_2}$这两个偏微分的值，这样就计算完了，这个步骤，就叫做Backward pass

在做Backward pass的时候，实际上的做法就是建另外一个neural network，本来正向neural network里面的activation function都是sigmoid function，而现在计算Backward pass的时候，就是建一个反向的neural network，它的activation function就是一个运算放大器op-amp，每一个反向neuron的input是loss $l$对后面一层layer的$z$的偏微分$\frac{\partial l}{\partial z}$，output则是loss $l$对这个neuron的$z$的偏微分$\frac{\partial l}{\partial z}$，做Backward pass就是通过这样一个反向neural network的运算，把loss $l$对每一个neuron的$z$的偏微分$\frac{\partial l}{\partial z}$都给算出来

注：如果是正向做Backward pass的话，实际上每次计算一个$\frac{\partial l}{\partial z}$，就需要把该neuron后面所有的$\frac{\partial l}{\partial z}$都给计算一遍，会造成很多不必要的重复运算，如果写成code的形式，就相当于调用了很多次重复的函数；而如果是反向做Backward pass，实际上就是把这些调用函数的过程都变成调用“值”的过程，因此可以直接计算出结果，而不需要占用过多的堆栈空间

#### Summary

最后，我们来总结一下Backpropagation是怎么做的

**Forward pass**，每个neuron的activation function的output，就是它所连接的weight的$\frac{\partial z}{\partial w}$

**Backward pass**，建一个与原来方向相反的neural network，它的三角形neuron的output就是$\frac{\partial l}{\partial z}$

把通过forward pass得到的$\frac{\partial z}{\partial w}$和通过backward pass得到的$\frac{\partial l}{\partial z}$乘起来就可以得到$l$对$w$的偏微分$\frac{\partial l}{\partial w}$
$$
\frac{\partial l}{\partial w} = \frac{\partial z}{\partial w}|_{forward\ pass} \cdot \frac{\partial l}{\partial z}|_{backward \ pass}
$$

<center><img src="https://gitee.com/Sakura-gh/ML-notes/raw/master/img/bp-summary.png" width="50%;"></center>
                
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    <div class="modal-content">
        <div class="search-header">
            <span class="title"><i class="fas fa-search"></i>&nbsp;&nbsp;搜索</span>
            <input type="search" id="searchInput" name="s" placeholder="请输入搜索的关键字"
                   class="search-input">
        </div>
        <div id="searchResult"></div>
    </div>
</div>

<script type="text/javascript">
$(function () {
    var searchFunc = function (path, search_id, content_id) {
        'use strict';
        $.ajax({
            url: path,
            dataType: "xml",
            success: function (xmlResponse) {
                // get the contents from search data
                var datas = $("entry", xmlResponse).map(function () {
                    return {
                        title: $("title", this).text(),
                        content: $("content", this).text(),
                        url: $("url", this).text()
                    };
                }).get();
                var $input = document.getElementById(search_id);
                var $resultContent = document.getElementById(content_id);
                $input.addEventListener('input', function () {
                    var str = '<ul class=\"search-result-list\">';
                    var keywords = this.value.trim().toLowerCase().split(/[\s\-]+/);
                    $resultContent.innerHTML = "";
                    if (this.value.trim().length <= 0) {
                        return;
                    }
                    // perform local searching
                    datas.forEach(function (data) {
                        var isMatch = true;
                        var data_title = data.title.trim().toLowerCase();
                        var data_content = data.content.trim().replace(/<[^>]+>/g, "").toLowerCase();
                        var data_url = data.url;
                        data_url = data_url.indexOf('/') === 0 ? data.url : '/' + data_url;
                        var index_title = -1;
                        var index_content = -1;
                        var first_occur = -1;
                        // only match artiles with not empty titles and contents
                        if (data_title !== '' && data_content !== '') {
                            keywords.forEach(function (keyword, i) {
                                index_title = data_title.indexOf(keyword);
                                index_content = data_content.indexOf(keyword);
                                if (index_title < 0 && index_content < 0) {
                                    isMatch = false;
                                } else {
                                    if (index_content < 0) {
                                        index_content = 0;
                                    }
                                    if (i === 0) {
                                        first_occur = index_content;
                                    }
                                }
                            });
                        }
                        // show search results
                        if (isMatch) {
                            str += "<li><a href='" + data_url + "' class='search-result-title'>" + data_title + "</a>";
                            var content = data.content.trim().replace(/<[^>]+>/g, "");
                            if (first_occur >= 0) {
                                // cut out 100 characters
                                var start = first_occur - 20;
                                var end = first_occur + 80;
                                if (start < 0) {
                                    start = 0;
                                }
                                if (start === 0) {
                                    end = 100;
                                }
                                if (end > content.length) {
                                    end = content.length;
                                }
                                var match_content = content.substr(start, end);
                                // highlight all keywords
                                keywords.forEach(function (keyword) {
                                    var regS = new RegExp(keyword, "gi");
                                    match_content = match_content.replace(regS, "<em class=\"search-keyword\">" + keyword + "</em>");
                                });

                                str += "<p class=\"search-result\">" + match_content + "...</p>"
                            }
                            str += "</li>";
                        }
                    });
                    str += "</ul>";
                    $resultContent.innerHTML = str;
                });
            }
        });
    };

    searchFunc('/search.xml', 'searchInput', 'searchResult');
});
</script>

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